An airline has81% of its flights depart on schedule. It has 69% of its flights depart and arrive on schedule. Find the probability that a flight that departs on schedule also arrives on schedule. Round the answer to two decimal places.
Let A be the event that a flight departs on schedule and B be the event that a flight arrives on schedule. We want to find P(B|A), the probability that a flight arrives on schedule given that it departs on schedule. Using Bayes' theorem:
P(B|A) = P(A|B)P(B)/P(A)
We know P(A) = 0.81, the probability that a flight departs on schedule. We also know P(B) = 0.69, the probability that a flight departs and arrives on schedule. To find P(A|B), the probability that a flight departs on schedule given that it arrives on schedule, we can rearrange the formula:
P(A|B) = P(B|A)P(A)/P(B)
We can solve for P(A|B) by plugging in the values we know:
P(A|B) = P(B|A)P(A)/P(B)
0.81 = P(B|A) * 0.81 / 0.69
P(B|A) = 0.9444
So the probability that a flight that departs on schedule also arrives on schedule is 0.94, rounded to two decimal places.
To find the probability that a flight that departs on schedule also arrives on schedule, we can use conditional probability.
Let's assume that A represents the event that a flight departs on schedule, and B represents the event that the same flight arrives on schedule.
The given probabilities are:
P(A) = 0.81 (81% of flights depart on schedule)
P(A ∩ B) = 0.69 (69% of flights depart and arrive on schedule)
We can use the formula for conditional probability:
P(B|A) = P(A ∩ B) / P(A)
Substituting the given values:
P(B|A) = 0.69 / 0.81
Now, let's calculate the probability:
P(B|A) = 0.69 / 0.81
P(B|A) ≈ 0.8529
Rounding the answer to two decimal places, the probability that a flight that departs on schedule also arrives on schedule is approximately 0.85.